What is the the vertex of #x = (y - 6)^2 - 11#?

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Answer 1 · 2016-01-16T22:17:07

#"Vertex" ->(x,y)->(-11,6)#

Given: #color(white)(....)x=(y-6)^2-11#............................(1)

View as the same thing as the vertex form for the U shaped quadratic but instead it is expressed in terms of #y# instead of #x#

Tony B

So the instead of stating that #x_("vertex")=(-1)xx(-6)# as in the U curve format we say #y_("vertex")=(-1)xx(-6)=6#

#y_("vertex")=6#

Substitute in equation (1) gives:

So #x_("vertex")=(6-6)^2-11=-11#

#"Vertex" ->(x,y)->(-11,6)#